1. Trigonometric
Ratios 1
A RATIO is a comparison
of two numbers. For
example;
boys to girls
cats : dogs
right : wrong.
In Trigonometry, the
comparison is between
sides of a triangle.
2. WEEK: 1
DATE: April 24th – April 28th, 2023
CLASS: Year 10
PERIODS: 1and 2
DURATION: 80 minutes.
TOPIC: Trigonometric Ratios 1
SUB-TOPIC: Trigonometric Ratios 1
INSTRUCTIONAL MATERIALS: Images of right- angled triangle.
REFERENCE BOOKS: Oluwaseun, A.J.S (2016).
WABP Essential Mathematics For
SS1. West African Book Publishers
Ltd. Pages:32 – 48.
PREVIOUS KNOWLEDGE: The students have been taught angles in a
triangle.
3. Performance Objectives
By the end of this lesson, the students should be able to:
• define sine, cosine and tangent of an acute angle.
• apply sine, cosine and tangent tables/ scientific
calculators to solve problems involving right- angled
triangles.
4. METHODOLOGY:
• Step 1: The teacher asks the students some questions from
the previous lesson.
• Step 2: With the aid of diagrams, teacher explains the
concept of trigonometry ratios to the students.
• Step3: The teacher guides the students to solve problems
on right – angled triangles using trigonometry ratios.
• Step 4: The students solve problems on right – angled
triangles using trigonometry ratios
• Step4: Teacher gives problems to the students to solve.
PEDAGOGICAL SKILLS: Discussion, Evaluation method,
Questioning method.
5. Trigonometric
Ratios 1
A RATIO is a comparison
of two numbers. For
example;
boys to girls
cats : dogs
right : wrong.
In Trigonometry, the
comparison is between
sides of a triangle.
6. We need to do some
housekeeping before we
can proceed…
7. In trigonometry, the ratio we are talking
about is the comparison of the sides of a
RIGHT TRIANGLE.
Two things MUST BE understood:
1. This is the hypotenuse.. This
will ALWAYS be the hypotenuse
2. This is 90°… this makes the
right triangle a right triangle…. Without
it, we can not do this trig… we WILL NOT
use it in our calculations because we
COULD NOT do calculations without it.
8. Now that we agree about the hypotenuse and right angle, there are only
4 things left; the 2 other angles and the 2 other sides.
A We will refer to the sides
in terms of their proximity
to the angle
If we look at angle A, there is
one side that is adjacent to it
and the other side is opposite
from it, and of course we have
the hypotenuse.
opposite
adjacent
hypotenuse
9. B
If we look at angle B, there is
one side that is adjacent to it
and the other side is opposite
from it, and of course we have
the hypotenuse.
opposite
adjacent
hypotenuse
11. θ this is the symbol for an unknown
angle measure.
It’s name is ‘Theta’.
Don’t let it scare you… it’s like ‘x’ except
for angle measure… it’s a way for us to
keep our variables understandable and
organized.
One more thing…
13. Trigonometric Ratios
Name
“say”
Sine Cosine tangent
Abbreviation
Abbrev.
Sin Cos Tan
Ratio of an
angle
measure
Sinθ = opposite side
hypotenuse
cosθ = adjacent side
hypotenuse
tanθ =opposite side
adjacent side
14. One more
time…
Here are the
ratios:
sinθ = opposite side
hypotenuse
cosθ = adjacent side
hypotenuse
tanθ =opposite side
adjacent side
15. In Geometry we learned three trigonometric ratios
made from the sides of a right triangle. They are
sine, cosine, and tangent.
We had an acronym to remember the ratios……
Soh Cah Toa……
𝛉
A
B
C
Opposite
Adjacent
16. Next, we will learn three additional trigonometric ratios.
They are the reciprocals of sine, cosine, and tangent.
Θ
A
B
C
Opposite
Adjacent
17. Trigonometry
03 September 2023
Learning Objective:
To be able to describe the sides of right-angled
triangle for use in trigonometry.
Trigonometry is concerned with the
connection between the sides and
angles in any right angled triangle.
Angle
18. A
A
The sides of a right -angled triangle are
given special names:
The hypotenuse, the opposite and the
adjacent.
The hypotenuse is the longest side and is
always opposite the right angle.
The opposite and adjacent sides refer to
another angle, other than the 90o.
19. There are three formulae involved in
trigonometry:
sin A=
cos A=
tan A =
S O H C A H T O
A
20. Using trigonometry on the calculator
All individual angles have different sine, cosine
and tangent ratios (or decimal values).
Scientific calculators store information about
every angle.
We need to be able to access this
information in the correct manner.
03 September
2023
Learning Objective:
To be able to use a scientific calculator to find
decimal values and angles in trigonometry.
21. Finding the ratios
The simplest form of question is finding the
decimal value of the ratio of a given angle.
Find:
1) sin 32 =
sin 32 =
2) cos 23 =
3) tan 78 =
4) tan 27 =
5) sin 68 =
22. Using ratios to find angles
We have just found that a scientific
calculator holds the ratio information
for sine (sin), cosine (cos) and
tangent (tan) for all angles.
It can also be used in reverse, finding
an angle from a ratio.
To do this we use the sin-1, cos-1 and
tan-1 function keys.
23. Example:
1. sin x = 0.1115 find angle x.
x = sin-1 (0.1115)
x = 6.4o
2. cos x = 0.8988 find angle x
x = cos-1 (0.8988)
x = 26o
sin-1 0.1115 =
shift sin
( )
cos-1 0.8988 =
shift cos
( )
25. Finding an angle from a triangle
To find a missing angle from a right-angled
triangle we need to know two of the sides of
the triangle.
We can then choose the appropriate ratio,
sin, cos or tan and use the calculator to
identify the angle from the decimal value of
the ratio.
Find angle C
a) Identify/label the
names of the sides.
b) Choose the ratio that
contains BOTH of the
letters.
14 cm
6 cm
C
1.
26. C = cos-1 (0.4286)
C = 64.6o
14 cm
6 cm
C
1.
H
A
We have been given
the adjacent and
hypotenuse so we use
COSINE:
Cos A =
hypotenuse
adjacent
Cos A =
h
a
Cos C =
14
6
Cos C = 0.4286
27. Find angle x
2.
8 cm
3 cm
x
A
O
Given adj and opp
need to use tan:
Tan A = adjacent
opposite
x = tan-1 (2.6667)
x = 69.4o
Tan A =
a
o
Tan x =
3
8
Tan x = 2.6667
28. 3.
12 cm
10 cm
y
Given opp and hyp
need to use sin:
Sin A = hypotenuse
opposite
x = sin-1 (0.8333)
x = 56.4o
sin A =
h
o
sin x =
12
10
sin x = 0.8333
30. Finding a side from a triangle
To find a missing side from a right-angled
triangle we need to know one angle and one
other side.
Cos45 =
13
x
To leave x on its own we need to
move the ÷ 13.
It becomes a “times” when it moves.
Note: If
Cos45 x 13 = x
31. Cos 30 x 7 = k
6.1 cm = k
7 cm
k
30o
1.
H
A
We have been given
the adj and hyp so we
use COSINE:
Cos A =
hypotenuse
adjacent
Cos A =
h
a
Cos 30 =
7
k
32. Tan 50 x 4 = r
4.8 cm = r
4 cm
r
50o
2.
A
O
Tan A =
a
o
Tan 50 =
4
r
We have been given
the opp and adj so we
use TAN:
Tan A =
33. Sin 25 x 12 = k
5.1 cm = k
12 cm
k
25o
3.
H
O
sin A =
h
o
sin 25 =
12
k
We have been given
the opp and hyp so we
use SINE:
Sin A =
34. x =
x
5 cm
30o
1.
H
A
Cos A =
h
a
Cos 30 =
x
5
30
cos
5
m
8 cm
25o
2.
H
O
m =
sin A =
h
o
sin 25 =
m
8
sin25
8
x = 5.8 cm
m = 18.9 cm
36. x =
x
5 cm
30o
1.
H
A
Cos A =
h
a
Cos 30 =
x
5
30
cos
5
x = 5.8 cm
4 cm
r
50o
2.
A
O
Tan 50 x 4 = r
4.8 cm = r
Tan A =
a
o
Tan 50 =
4
r
3.
12 cm
10 cm
y
y = sin-1 (0.8333)
sin A =
h
o
sin y =
12
10
sin y = 0.8333
3.
12 cm
10 cm
y
y = sin-1 (0.8333)
y = 56.4o
sin A =
h
o
sin y =
12
10
sin y = 0.8333
37. x =
x
5 cm
30o
1.
H
A
Cos A =
h
a
Cos 30 =
x
5
30
cos
5
x = 5.8 cm
4 cm
r
50o
2.
A
O
Tan 50 x 4 = r
4.8 cm = r
Tan A =
a
o
Tan 50 =
4
r
3.
12 cm
10 cm
y
y = sin-1 (0.8333)
y = 56.4o
sin A =
h
o
sin y =
12
10
sin y = 0.8333
41. Applications of Trigonometry Ratios
W.E.1:
The perimeter of an isosceles triangle is 32.8 cm. If the base of
the triangle is 12.8 cm. Find (a) its vertical angle (b) its
altitude.
Solution
42. W.E.2:
A ladder of length 4.5 m leans against a vertical wall
making an angle of 50° with the horizontal . If the bottom of
the window is 4m above the ground, what is the distance
between the top of the ladder and the bottom of the
window? (Give your answer correct to the nearest cm)
Solution
44. EVALUATION
1. A ladder of length 6.0 m rests with its foot on a horizontal
ground and leans against a vertical wall. The inclination
of
the ladder to the horizontal is 80°. Find correct to one
decimal place.
(a) the distance of the foot of the ladder from the wall.
(b) the height above the ground at which the upper
end of the ladder touches the wall.
46. WEEK: 1
DATE: April 24th – April 28th, 2023
CLASS: Year 10
PERIODS: 3 and 4
DURATION: 80 minutes.
TOPIC: Trigonometric Ratios 1
SUB-TOPIC: Complementary Angles.
INSTRUCTIONAL MATERIALS: Images of right- angled triangle.
REFERENCE BOOKS: Oluwaseun, A.J.S (2016).
WABP Essential Mathematics For
SS1. West African Book Publishers
Ltd. Pages:32 – 48.
PREVIOUS KNOWLEDGE: The students have been taught angles in a
triangle.
47. Performance Objectives
By the end of this lesson, the students will be able to:
• State complementary angle formula and apply the
formula to solve problems on right-angled triangle.
48. METHODOLOGY:
• Step 1: The teacher asks the students some questions from
the previous lesson.
• Step 2: With the aid of diagrams, teacher explains the
concept of complementary angles to the students.
• Step3: The teacher guides the students to solve problems
on right – angled triangles using complementary angle
formula.
• Step 4: The students solve problems on right – angled
triangles using complementary angle formula.
• Step4: Teacher gives problems to the students to solve.
PEDAGOGICAL SKILLS: Discussion, Evaluation method,
Questioning method.
49. Complementary angles
Two angles are said to be complementary when they add
up to 90°.
For example, in ∆𝐴𝐵𝐶, below, angle C and A are
complementary.
55. WEEK: 1
DATE: April 24th – April 28th, 2023
CLASS: Year 10
PERIODS: 5
DURATION: 40 minutes.
TOPIC: Trigonometric Ratios 1
SUB-TOPIC: Length of chord using trigonometry ratios.
INSTRUCTIONAL MATERIALS: Images of right- angled triangle.
REFERENCE BOOKS: Oluwaseun, A.J.S (2016).
WABP Essential Mathematics For
SS1. West African Book Publishers
Ltd. Pages:32 – 48.
PREVIOUS KNOWLEDGE: The students have been taught angles in a
triangle.
56. Performance Objectives
By the end of this lesson, the students should be able to:
• determine the length of chord using trigonometry ratios.
57. METHODOLOGY:
• Step 1: The teacher asks the students some questions from
the previous lesson.
• Step 2: With the aid of diagrams, teacher explains the
concept of length of chord to the students.
• Step3: The teacher guides the students to solve problems
on length of chord using Pythagoras’ theorem.
• Step 4: The students solve problems on length of chord
using Pythagoras’ theorem.
• Step4: Teacher gives problems to the students to solve.
PEDAGOGICAL SKILLS: Discussion, Evaluation method,
Questioning method.
58. Determine the length of chord using trigonometry ratios
Trigonometric ratios can be used to find the length of chords of a given
circle. However, in some cases where angles are not given, Pythagoras
theorem is used to find the lengths of chords in such cases.
Pythagoras theorem is stated as follows:
It states that c2 = a2 + b2
Pythagoras theorem states that in a right angled triangle, the square
of the length of the hypotenuse is equal to the sum of the square of the
lengths of the other two sides.
59. W.E.1:
A chord is drawn 3cm away from the centre of a circle of
radius 5cm. Calculate the length of the chord.
Solution
From the diagram above in right –angled triangle ABO.
𝐴𝐵2 + 𝐵𝑂2 = 𝐴𝑂2 𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑎𝑠′𝑡ℎ𝑒𝑜𝑟𝑒𝑚
𝐴𝐵2 = 𝐴𝑂2- 𝐵𝑂2
𝐴𝐵2 = 52 − 32
= 25 – 9
AB = 16
AB = 4 cm
Since AB is the midpoint of chord AC, then
Length of chord AC = 2× 𝐴𝐵
= 2 × 4 𝑐𝑚 = 8 𝑐𝑚
B
60. W.E.2:
In the figure below, O is the centre of circle, HKL. HK =
16cm, HL = 10cm and the perpendicular from O to the HK is
4cm. What is the length of the perpendicular from O to HL?
Solution
Let the distance from O to HL= x cm
In right-angled triangle OMH:
61. Applying pythagoras’ theorem
𝑂𝐻2
= 𝐻𝑀2
+ 𝑀𝑂2
𝑂𝐻2 = 82 + 42
𝑂𝐻2 = 64 + 16
= 80
OH = √80
But |OH| = |OL| =√80 (radius of the circle )
In right-angled triangle ONL
𝑂𝐿2 = 𝑂𝑁2 + 𝑁𝐿2
√80
2
= 𝑥2 + 52
𝑥2 = 80 − 25
x = 55 𝑐𝑚
x = 7. 416 cm
The length of the perpendicular from O to HL = 7.416 cm.
62. EVALUATION 3
1.A chord 30cm long is 20cm from the centre of a circle.
Calculate the length of the chord which is 24cm from the
centre .
2. Given the figure below, calculate the length of the
chord AB..
r =14cm
O
r
B
A
Where r = 14 cm, < AOB = 58°
63. ASSIGNMENT 3
1. A chord PQ of a circle is 12.4 cm long. The radius of the
circle is 8.5 cm. Calculate the distance of the mid-
point
of the chord PQ from the centre, O of the circle . Give
your answer to 3 s.f.
2. AB is a chord of a circle. The radius of the circle is 16
cm
and the distance of the mid –point of the chord from the
centre of the circle O, is 10 cm. Calculate to 1 d.p
(a) the length of the chord AB
(b) the angle subtends at the centre of the circle by
chord AB
